Coulomb's law — force, comparison with gravity: Worked examples

1.8.2 · D3 · Physics › Electromagnetism › Coulomb's law — force, comparison with gravity

Scenario matrix

Jo kuch bhi Coulomb's law exam mein de sakta hai, woh in cells mein se kisi ek mein aata hai. Hum har ek ko cover karenge.

#	Cell (case class)	Tricky kya hai	Example
A	Do charges, opposite signs	direction = attraction	Ex 1
B	Do charges, like signs	direction = repulsion	Ex 2
C	Scaling ( $r$ ya $q$ change karo)	inverse-square vs linear	Ex 3
D	Ek charge zero hai (degenerate)	force zero honi chahiye	Ex 4
D′	Coincident charges $r \to 0$ (forbidden/limit)	formula blow up karta hai; domain	Ex 4b
E	Teen charges ek line mein (vector add, 1-D)	signs left/right set karte hain	Ex 5
F	Non-collinear, sab like signs (2-D)	components + symmetry	Ex 6
F′	Non-collinear, mixed signs (2-D)	ek push, ek pull	Ex 6b
G	Equilibrium point (jahan net force = 0)	limiting / algebra	Ex 7
H	Real-world word problem	words → numbers translate karo	Ex 8
I	Exam twist (gravity se compare / ratios)	conceptual, $r$ cancel ho jaata hai	Ex 9

Shuru karne se pehle, ek reminder us tool ka jo hum baar baar use karenge:

Cell A — opposite signs (attraction)

Worked example Example 1 — ek plus aur ek minus

Charges $q_{1} = + 4 μ C$ aur $q_{2} = - 5 μ C$ ka separation $r = 0.20 m$ hai. Force ki magnitude aur direction nikalo.

Forecast: signs opposite hain, isliye koi bhi arithmetic karne se pehle — kya woh ek doosre ko kheenchenge ya door dhakeleinge? (Guess: kheenchenge.)

Step 1 — SI mein convert karo, aur sign deliberately handle karo. $q_{1} = + 4 μ C = 4 \times 1 0^{- 6} C$ aur $q_{2} = - 5 μ C = - 5 \times 1 0^{- 6} C$ . Magnitude formula ko $∣ q_{1} q_{2} ∣$ chahiye, isliye hum sirf sizes lete hain: $∣ q_{1} ∣ = 4 \times 1 0^{- 6}$ , $∣ q_{2} ∣ = 5 \times 1 0^{- 6}$ , $r = 0.20 m$ . Ye step kyun, aur minus sign drop karna allowed kyun hai? µC → C convert karna sirf ek scale change hai ( $\times 1 0^{- 6}$ ); ye sign ko kabhi nahi chhuta. Hum $q_{2}$ ka minus isliye nahi drop karte ki woh matter nahi karta, balki isliye ki humne kaam ko do halves mein baant diya hai: magnitude formula poochta hai "kitna bada?", aur iske construction mein woh sirf $∣ q_{1} q_{2} ∣$ use karta hai, isliye negative number ki usme jagah hi nahi. Minus sign phenka nahi gaya — usse park kiya gaya hai aur Step 4 mein direction decide karne ke liye wapas laaya jaayega. Ye wahi sign-vs-magnitude discipline hai jo parent note insist karta hai.

Step 2 — Magnitude formula, $∣ q_{1} q_{2} ∣$ use karte hue. $F = (8.99 \times 1 0^{9}) \frac{( 4 \times 1 0 ^{- 6} ) ( 5 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}}$ Yahan magnitudes kyun? Hum pehle size calculate karte hain; sign kabhi magnitude formula mein belong nahi karta.

Step 3 — Calculate karo. Andar numerator: $(4 \times 1 0^{- 6}) (5 \times 1 0^{- 6}) = 20 \times 1 0^{- 12}$ . $(0.20)^{2} = 0.04$ se divide karo: $20 \times 1 0^{- 12} /0.04 = 5 \times 1 0^{- 10}$ . $k$ se multiply karo: $8.99 \times 1 0^{9} \times 5 \times 1 0^{- 10} = 4.495 N$ . Kyun in teen sub-steps mein todte hain? Powers of ten wahan hain jahan sign-and-exponent slips hote hain; charge product, phir distance square, phir $k$ se multiply — in alag groups mein rakhne se har exponent visible aur checkable rehta hai, bajaye ek saath sab juggle karne ke.

Step 4 — Parked sign wapas laao direction ke liye. $q_{1} q_{2} = (+) (-) < 0 \Rightarrow$ attractive. Forecast sahi nikla.

Answer: $F \approx 4.50 N$ , dono charges ek doosre ki taraf kheench rahe hain.

Verify: Units: $\frac{N\cdotm ^{2}}{C ^{2}} \cdot \frac{C ^{2}}{m ^{2}} = N$ ✓. $20 cm$ par microcoulombs ke liye kuch newtons ka magnitude — sahi ballpark hai.

Cell B — like signs (repulsion)

Worked example Example 2 — do positives

Ab $q_{1} = + 4 μ C$ , $q_{2} = + 5 μ C$ , same $r = 0.20 m$ . Force?

Forecast: numbers Ex 1 jaisi hi hain (same magnitudes, same distance). Toh kya force ki size change hogi? (Guess: nahi — sirf direction.)

Step 1 — Notice karo ki magnitudes unchanged hain. $∣ q_{1} q_{2} ∣$ aur $r$ exactly pehle jaisi hain. Ye step kyun? Magnitude formula sirf $∣ q_{1} q_{2} ∣$ dekhta hai; sign $+$ mein flip karne se woh nahi badalta.

Step 2 — Toh $F = 4.50 N$ phir se, dubara compute karne ki zaroorat nahi.

Step 3 — Signs se direction. $q_{1} q_{2} = (+) (+) > 0 \Rightarrow$ repulsive, door dhakelta hai.

Answer: $F \approx 4.50 N$ , charges ko door dhakelta hai.

Verify: Same numbers in, same magnitude out ✓. Ye prove karta hai ki magnitude sign ke liye blind hai; sirf word (attract/repel) badla. Directly Ex 1 se compare karo.

Cell C — scaling (inverse-square)

Worked example Example 3 — distance aur charge dono change karo

Ex 1 se shuru karo ( $F = 4.50 N$ ). Ab distance triple karo aur $q_{1}$ double karo. Naya force?

Forecast: charge double karna $F$ ko $2$ se multiply karta hai; $r$ triple karna $F$ ko... $3$ se divide karta hai? ya $9$ se? (Padhne se pehle guess karo — inverse-square wahi trap hai.)

Step 1 — $F$ ko ratios ke product ke roop mein likho. Kyunki $F = k \frac{∣ q _{1} q _{2} ∣}{r ^{2}}$ hai, $q_{1}$ ko factor $a$ se aur $r$ ko factor $b$ se scale karne par $F_{new} = F_{old} \cdot \frac{a}{b ^{2}} .$ Ye step kyun? Ratios se hum purana answer reuse kar sakte hain scratch se recompute kiye bina — aur woh exactly dikhate hain ki square kahan hai.

Step 2 — $a = 2$ (double $q_{1}$ ), $b = 3$ (triple $r$ ) daalo. $F_{new} = 4.50 \times \frac{2}{3 ^{2}} = 4.50 \times \frac{2}{9} .$ $3^{2}$ kyun, $3$ kyun nahi? Distance $r^{2}$ ke roop mein aata hai, isliye uska factor squared hota hai. Ye is cell ka pura point hai.

Step 3 — Evaluate karo. $4.50 \times 2/9 = 4.50 \times 0.2222 = 1.00 N$ .

Answer: $F_{new} \approx 1.00 N$ .

Verify: $\frac{2}{9} \approx 0.222$ ; $4.50 \times 0.222 = 1.00$ ✓. Agar tumne galti se $\frac{2}{3}$ use kiya hota toh $3.0 N$ milta — teen guna zyada. Isliye squared distance matter karta hai.

Cell D — zero charge (degenerate case)

Worked example Example 4 — agar ek charge zero ho toh?

$q_{1} = + 7 μ C$ , $q_{2} = 0$ , distance $r = 0.05 m$ . Force?

Forecast: koi charge nahi wala object — kya Coulomb's law apply bhi hota hai? $F$ kya honi chahiye? (Guess: exactly zero.)

Step 1 — $q_{2} = 0$ substitute karo. $F = k \frac{∣ q _{1} \cdot 0∣}{r ^{2}} = k \frac{0}{r ^{2}} = 0.$ Ye step kyun? Product $q_{1} q_{2}$ mein zero factor hai, jo pura numerator kill kar deta hai, chahe $q_{1}$ ya $r$ kuch bhi ho.

Step 2 — Interpret karo. Koi charge nahi matlab $q_{1}$ ke liye pakadne ko kuch nahi. Force identically zero hai, aur uski direction undefined hai (zero vector ki koi direction nahi hoti). Direction mention kyun? Taaki tum "attract ya repel?" poochh kar time waste na karo — nahi kuchh; force hai hi nahi.

Answer: $F = 0 N$ , kisi bhi distance $r > 0$ ke liye.

Verify: $∣ q_{1} \cdot 0∣ = 0$ , isliye $F = 0$ ✓. Limiting behaviour: jaise $q_{2} \to 0$ , $F \to 0$ continuously — koi surprise nahi.

Cell D′ — coincident charges, $r \to 0$ (forbidden input / limit)

Worked example Example 4b — charges ke touch karne par kya hota hai?

$q_{1} = + 7 μ C$ , $q_{2} = + 2 μ C$ rakho lekin gap shrink karo: $r = 0.10$ , phir $0.010$ , phir $0.0010 m$ par $F$ compute karo, aur poochho ki $r = 0$ kya dega.

Forecast: jaise $r$ chhota hota hai, kya $F$ kisi finite value par settle ho jaata hai, ya bound ke bina badhta hai? (Compute karne se pehle guess karo.)

Step 1 — Charge product fixed hai. $k ∣ q_{1} q_{2} ∣ = (8.99 \times 1 0^{9}) (7 \times 1 0^{- 6}) (2 \times 1 0^{- 6}) = 0.1259 N\cdotm^{2}$ (yeh poora lump yahan constant hai). Isse isolate kyun karo? Sirf $r$ change ho raha hai, isliye constant bahar kheenchne se clearly dikhta hai ki $F = 0.1259/ r^{2}$ hai.

Step 2 — Shrinking $r$ par evaluate karo. $r = 0.10 : F = \frac{0.1259}{0.01} = 12.6 N; r = 0.010 : F = \frac{0.1259}{1 0 ^{- 4}} = 1.26 \times 1 0^{3} N; r = 0.0010 : F = \frac{0.1259}{1 0 ^{- 6}} = 1.26 \times 1 0^{5} N .$ Sequence kyun? $r$ ka har das guna shrink $F$ ko $100$ se multiply karta hai ( $r^{2}$ ki wajah se). Pattern trend ko unmistakable bana deta hai.

Step 3 — Limit aur forbidden point. Jaise $r \to 0$ , $F = 0.1259/ r^{2} \to + \infty$ : force diverge karta hai. Exactly $r = 0$ par expression $1/ r^{2}$ undefined hai (zero se division), isliye Coulomb's law koi answer hi nahi deta — $r = 0$ uski domain ke bahar hai, jaise upar bataya. Yeh physics ko break kyun nahi karta? Do ideal point charges ek hi point par ek mathematical fiction hai; real charged objects ka size hota hai aur woh ek point occupy nahi kar sakte, aur tiny separations par doosri physics (quantum effects, nucleus mein strong force) over le leti hai. Coulomb's law ek large-scale, point-charge law hai aur simply $r = 0$ par apply nahi hota.

Answer: $F \to \infty$ jaise $r \to 0$ ; at $r = 0$ value undefined hai (forbidden input).

Verify: $0.1259/0.01 = 12.59$ ; $0.1259/1 0^{- 4} = 1259$ ; har step $\times 100$ ✓ — genuinely blow-up ho raha hai, singularity confirm.

Cell E — teen charges ek line mein (1-D superposition)

Worked example Example 5 — ek charge do doosron ko feel karta hai

Teen charges ek straight line par (positions metres mein): $q_{A} = + 2 μ C$ at $x = 0$ , $q_{B} = + 3 μ C$ at $x = 0.30$ , aur humara target $q_{C} = + 1 μ C$ at $x = 0.10$ . $q_{C}$ par net force nikalo.

Figure (Cell E). Horizontal axis woh line hai jis par teenon charges baithe hain. Blue dot $q_{A}$ hai, pink dot $q_{B}$ hai, beech mein yellow dot humara target $q_{C}$ hai. $q_{C}$ se nikalne wale do coloured arrows do pushes hain: lamba blue arrow (paas wale charge $q_{A}$ se) right point karta hai; chhota pink arrow (door wale charge $q_{B}$ se) left point karta hai. Picture ka point yeh hai ki length = strength — tum literally dekh sakte ho ki blue arrow pink ko outreach karta hai, isliye kisi bhi arithmetic se pehle diagram batata hai ki net force rightward hai.

Forecast: $q_{C}$ do positive charges ke beech baitha hai. $q_{A}$ (left) use right push karta hai; $q_{B}$ (right) use left push karta hai. Kaun jeetega — aur toh $q_{C}$ kis taraf move karega? (Compute karne se pehle guess karo.)

Step 1 — $q_{A}$ se $q_{C}$ par force. Separation $r_{A C} = 0.10 m$ . $F_{A C} = (8.99 \times 1 0^{9}) \frac{( 2 \times 1 0 ^{- 6} ) ( 1 \times 1 0 ^{- 6} )}{( 0.10 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{2 \times 1 0 ^{- 12}}{0.01} = 1.798 N .$ Dono positive → repulsion → yeh $q_{C}$ ko $+ x$ direction mein push karta hai (right). Ye step kyun? Superposition kehta hai har source ko akele treat karo — maan lo $q_{B}$ hai hi nahi — pehle plain two-charge force compute karo. $q_{B}$ ko baad mein fold karenge; unhe ek ek karke karna wahi cheez hai jo signs ko tangle hone se rokti hai.

Step 2 — $q_{B}$ se $q_{C}$ par force. Separation $r_{B C} = 0.30 - 0.10 = 0.20 m$ . $F_{B C} = (8.99 \times 1 0^{9}) \frac{( 3 \times 1 0 ^{- 6} ) ( 1 \times 1 0 ^{- 6} )}{( 0.20 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{3 \times 1 0 ^{- 12}}{0.04} = 0.674 N .$ Dono positive → repulsion → yeh $q_{C}$ ko $- x$ direction mein push karta hai (left). Ye step kyun? Step 1 jaisa hi routine, lekin doosre source ke liye. Notice karo $q_{B}$ bada hai ( $3$ vs $2 μ C$ ) phir bhi do guna door hai; $1/ r^{2}$ penalty wahi cheez hai jo hum raw charge advantage ke against test kar rahe hain.

Step 3 — Signed 1-D vectors ke roop mein add karo (right = $+$ lo). $F_{net} = + 1.798 - 0.674 = + 1.124 N .$ Subtract kyun? Dono pushes ek hi line par opposite directions mein point karte hain, isliye unke signed values add hote hain — subtraction ek tarah se disguise mein hai.

Answer: $F_{net} \approx 1.12 N$ right ( $+ x$ ) direction mein. Paas wala charge $q_{A}$ jeetta hai.

Verify: $1.798 - 0.674 = 1.124$ ✓. Sanity: $q_{A}$ paas hai ( $0.10$ vs $0.20$ ), aur paas hona $1/ r^{2}$ ki tarah matter karta hai, isliye woh dominate karta hai chahe $q_{B}$ bada ho — exactly wahi jo figure mein arrows dikhate hain.

Cell F — non-collinear, sab like signs (2-D vector addition)

Worked example Example 6 — right-triangle setup

$q_{1} = + 6 μ C$ origin $(0, 0)$ par, aur $q_{2} = + 6 μ C$ $(0.40, 0) m$ par. Test charge $q_{3} = + 2 μ C$ midpoint ke upar $(0.20, 0.15) m$ par baitha hai. $q_{3}$ par net force nikalo.

Figure (Cell F). Do neeche wale dots ( $q_{1}$ blue, $q_{2}$ pink) aur utha hua dot ( $q_{3}$ yellow) ek isosceles triangle banaate hain. Har dashed line $0.25 m$ leg hai. Blue aur pink arrows $q_{3}$ par do pushes hain, har ek apne source se seedha door point karta hai; woh symmetrically baahir jhukti hain. Yellow arrow unka sum hai, ekdum vertical khada hai. Picture cancellation visible banata hai: do slanted arrows mirror images hain, isliye unke sideways parts opposite directions mein point karte hain aur delete ho jaate hain, jabki unke upward parts stack hote hain — isliye net arrow purely vertical hai.

Forecast: symmetry se $q_{1}$ aur $q_{2}$ $q_{3}$ se guzarne wali vertical ke baare mein mirror images hain. Jab hum unke do pushes add karte hain, kaun se components cancel hote hain aur kaun se reinforce karte hain? (Guess: horizontals cancel, verticals add — net force straight up.)

Step 1 — Har source se $q_{3}$ tak distances. $q_{1} \to q_{3}$ : $Δ = (0.20, 0.15)$ , isliye $r_{1} = 0.2 0^{2} + 0.1 5^{2} = 0.0625 = 0.25 m$ . Symmetry se $r_{2} = 0.25 m$ bhi. Ye step kyun? $1/ r^{2}$ use karne se pehle $r$ chahiye; geometry $0.05$ se scaled $3$ - $4$ - $5$ triangle hai.

Step 2 — Har force ki magnitude (symmetry se dono equal hain). $F_{1} = (8.99 \times 1 0^{9}) \frac{( 6 \times 1 0 ^{- 6} ) ( 2 \times 1 0 ^{- 6} )}{( 0.25 ) ^{2}} = 8.99 \times 1 0^{9} \times \frac{12 \times 1 0 ^{- 12}}{0.0625} = 1.726 N .$ Sirf ek kyun compute karo? Setup mirror image hai, isliye $F_{2}$ ka identical magnitude hai; ise do baar compute karna effort waste hoga. Hum unhe sirf direction mein alag karenge agले step mein.

Step 3 — Har force ko components mein todо. Teenon charges positive hain, isliye har force $q_{3}$ ko apne source se door push karti hai. $q_{1}$ se force $(0.20, 0.15) /0.25 = (0.8, 0.6)$ along point karti hai. $F_{1 x} = 1.726 \times 0.8 = 1.381, F_{1 y} = 1.726 \times 0.6 = 1.036 N .$ $q_{2}$ se force $(0.20 - 0.40, 0.15) /0.25 = (- 0.8, 0.6)$ along point karti hai: $F_{2 x} = - 1.381, F_{2 y} = + 1.036 N .$ Components kyun? Alag directions mein vectors numbers ki tarah add nahi ho sakte; hum har ek ko $x$ aur $y$ mein split karte hain, un alag alag add karte hain, phir recombine karte hain. Yeh single reliable 2-D method hai.

Step 4 — Components sum karo. $F_{x} = 1.381 + (- 1.381) = 0, F_{y} = 1.036 + 1.036 = 2.072 N .$ Forecast holds: horizontals cancel, verticals reinforce.

Answer: $F_{net} \approx 2.07 N$ seedha up ( $+ y$ ).

Verify: $F_{x} = 0$ exact symmetry se ✓; $2 \times 1.036 = 2.072$ ✓. Magnitude $0^{2} + 2.07 2^{2} = 2.072 N$ , $2 F_{1} = 3.45 N$ se kam hai, kyunki sirf vertical slices bachti hain — sensible.

Cell F′ — non-collinear, mixed signs (ek push, ek pull) 2-D mein

Worked example Example 6b — ek pull aur ek push vectors ke roop mein add karo

Ex 6 jaisi geometry, lekin ab $q_{2}$ ka sign flip karo: $q_{1} = + 6 μ C$ at $(0, 0)$ , $q_{2} = - 6 μ C$ at $(0.40, 0) m$ , test charge $q_{3} = + 2 μ C$ at $(0.20, 0.15) m$ . $q_{3}$ par net force nikalo.

Forecast: $q_{1}$ (like sign) $q_{3}$ ko door push karta hai; $q_{2}$ (opposite sign) $q_{3}$ ko apni taraf pull karta hai. Unki magnitudes ab bhi equal hain. Jab hum ek push aur ek pull add karte hain jo mirror images hain, kaun se components ab survive karte hain — vertical ya horizontal? (Compute karne se pehle guess karo — Ex 6 se flip hoga.)

Step 1 — Magnitudes unchanged. Distances ab bhi $0.25 m$ hain aur $∣ q_{1} q_{2} ∣ = ∣ q_{1} q_{3} ∣$ pairs ki sizes same hain, isliye har force magnitude phir se $F_{1} = F_{2} = 1.726 N$ hai. Unhe reuse kyun karo? Magnitude sirf $∣ q ∣$ aur $r$ par depend karta hai, dono unchanged; sirf directions differ karte hain. Yeh new physics (mixed signs) ko cleanly isolate karta hai.

Step 2 — $q_{1}$ se force ki direction (push, $q_{1}$ se door). Pehle jaisi hi: unit vector $(0.8, 0.6)$ . $F_{1 x} = + 1.381, F_{1 y} = + 1.036 N .$ Yeh direction kyun? $q_{1}$ aur $q_{3}$ dono positive hain → repulsion → force $q_{1}$ se $q_{3}$ ki taraf point karta hai, yaani up-and-to-the-right.

Step 3 — $q_{2}$ se force ki direction (pull, $q_{2}$ ki taraf). $q_{2}$ negative hai, $q_{3}$ positive → attraction → $q_{3}$ par force $q_{3}$ se $q_{2}$ ki taraf point karta hai, yaani $(0.40 - 0.20, 0 - 0.15) /0.25 = (0.8, - 0.6)$ along. $F_{2 x} = 1.726 \times 0.8 = + 1.381, F_{2 y} = 1.726 \times (- 0.6) = - 1.036 N .$ Sign sirf $y$ -part kyun flip karta hai? Attraction arrow ko Ex 6 ke relative reverse kar deta hai: ab woh neeche $q_{2}$ ki taraf aim karta hai instead of upar door. Horizontal parts line up ho jaate hain; verticals ab oppose karte hain.

Step 4 — Components sum karo. $F_{x} = 1.381 + 1.381 = 2.762, F_{y} = 1.036 + (- 1.036) = 0 N .$ Ex 6 se exactly ulta survivor: ab horizontals reinforce karte hain aur verticals cancel ho jaate hain.

Answer: $F_{net} \approx 2.76 N$ seedha right ( $+ x$ , $q_{1}$ ki side se $q_{2}$ ki side ki taraf) point karta hai.

Verify: $F_{y} = 0$ push-plus-pull ki mirror symmetry se ✓; $2 \times 1.381 = 2.762$ ✓. Physical sense: $q_{3}$ $+$ charge se door shove hota hai aur $-$ charge ki taraf pull hota hai — dono effects same horizontal direction mein point karte hain, isliye woh add hote hain. Yeh Electric Field ka classic "+ se − ki taraf field point karta hai" behaviour hai.

Cell G — equilibrium point (net force = 0)

Worked example Example 7 — test charge kahan kuch feel nahi karta?

Do fixed charges: $q_{1} = + 9 μ C$ at $x = 0$ aur $q_{2} = + 4 μ C$ at $x = 1.0 m$ . Line par kahan ek chhota charge zero net force ke saath baaith sakta hai?

Figure (Cell G). Blue dot ( $q_{1}$ , bada charge) left par hai, pink dot ( $q_{2}$ ) right par hai, aur yellow square balance point mark karta hai. Square se nikalne wale do chhote arrows opposite directions mein point karte hain aur equal length ke draw kiye gaye hain — woh equal length hi poori condition hai "net force = 0." Crucially yellow square midpoint se past hai, chhote pink charge ke paas: diagram dikhata hai kyun balance chhote charge ki taraf drift karna chahiye, taaki uski chhoti distance uske chhote charge ki compensate kare.

Forecast: dono fixed charges positive hain, isliye unke beech ek test charge har ek se door push hota hai. Equilibrium wahan hai jahan dono pushes balance karte hain. Kyunki $q_{1}$ bada hai, kya balance point bade charge ke paas hoga ya chhote ke? (Guess: chhote charge $q_{2}$ ke paas, taaki bade wale ki extra distance use weaken kare.)

Step 1 — Dono magnitudes equal set karo. Test charge distance $x$ par $q_{1}$ se ho, isliye $q_{2}$ se distance $(1 - x)$ hai. $k$ aur test charge $q$ dono sides par cancel ho jaate hain: $\frac{q _{1}}{x ^{2}} = \frac{q _{2}}{( 1 - x ) ^{2}} .$ Equal kyun set karo? "Zero net force" matlab leftward push rightward push ke magnitude ke equal hai — woh $x$ ke liye equation hai.

Step 2 — Charges daalo (µC mein — units ratio mein cancel ho jaate hain). $\frac{9}{x ^{2}} = \frac{4}{( 1 - x ) ^{2}} .$

Step 3 — Square root lo (dono sides positive hain, isliye yeh safe hai): $\frac{3}{x} = \frac{2}{1 - x} .$ Ab square root kyun? Yeh messy squared equation ko clean linear mein badal deta hai.

Step 4 — Cross-multiply aur solve karo. $3 (1 - x) = 2 x \Rightarrow 3 - 3 x = 2 x \Rightarrow 3 = 5 x \Rightarrow x = 0.6 m$ .

Answer: balance point $x = 0.60 m$ par hai $q_{1}$ se ( $q_{2}$ se $0.40 m$ ).

Verify: $\frac{9}{0. 6 ^{2}} = \frac{9}{0.36} = 25$ ; $\frac{4}{0. 4 ^{2}} = \frac{4}{0.16} = 25$ ✓. Aur $0.60 > 0.50$ matlab yeh hai chhote charge $q_{2}$ ke paas — forecast confirm. (Doosra algebraic root, $x = 3$ , segment ke bahar hai aur sign check se reject hoga — do like charges ke beech sirf andar wala root real balance hai.)

Cell H — ek real-world word problem

Worked example Example 8 — static-cling balloon

Tum ek balloon ko baalon par ragdate ho; woh $q = - 8 \times 1 0^{- 8} C$ charge pick up karta hai. Tum use wall ke ek chhote charged spot $Q = + 5 \times 1 0^{- 8} C$ se $d = 3.0 cm$ door laate ho. Kaun si force balloon ko wall se pakadti hai, aur kis direction mein?

Forecast: ragadne se balloon ko negative charge milta hai, spot positive hai — opposite signs. Attract ya repel? Aur kya force chhoti hogi (fraction of newton) ya badi? (Compute karne se pehle guess karo.)

Step 1 — Words ko SI mein translate karo, aur signs alag karo. $∣ q ∣ = 8 \times 1 0^{- 8} C$ , $∣ Q ∣ = 5 \times 1 0^{- 8} C$ , $d = 0.030 m$ . Actual signs baad ke liye alag record karo: $q$ $-$ hai, $Q$ $+$ hai. Ye step kyun, aur sign alag kyun karo? "3.0 cm" aur nanocoulomb-scale numbers metres aur coulombs ban ne chahiye $k$ use karne se pehle. Magnitude formula $F = k ∣ q Q ∣/ d^{2}$ sirf sizes leta hai — agar tum isme negative $q$ feed karo toh "negative force" milega, jo meaningless hai (force size negative nahi ho sakta). Isliye hum signs deliberately abhi side mein rakhte hain, size calculation karte hain, aur final step mein signs wapas laate hain direction decide karne ke liye. Yeh exactly wahi "sign vs magnitude" discipline hai jo Electric Field aur parent note mein hai.

Step 2 — Magnitude formula. $F = (8.99 \times 1 0^{9}) \frac{( 8 \times 1 0 ^{- 8} ) ( 5 \times 1 0 ^{- 8} )}{( 0.030 ) ^{2}} .$

Step 3 — Calculate karo, ek factor at a time. Numerator: $(8 \times 1 0^{- 8}) (5 \times 1 0^{- 8}) = 40 \times 1 0^{- 16} = 4.0 \times 1 0^{- 15}$ . $(0.030)^{2} = 9.0 \times 1 0^{- 4}$ se divide karo: $4.0 \times 1 0^{- 15} /9.0 \times 1 0^{- 4} = 4.44 \times 1 0^{- 15 + 4} = 4.44 \times 1 0^{- 12}$ . $k$ se multiply karo: $8.99 \times 1 0^{9} \times 4.44 \times 1 0^{- 12} = 3.99 \times 1 0^{- 2} N$ . Aise group kyun karo? Exponents ( $- 8$ do baar, phir $- 4$ , phir $+ 9$ ) danger zone hain; charge product, phir distance square, phir $k$ — alag alag moves mein karne se har power of ten view mein rehta hai taaki slip turant pakad mein aaye.

Step 4 — Direction ke liye signs wapas laao. $q$ negative hai aur $Q$ positive hai → opposite signs → attraction; balloon wall ki taraf pull hota hai.

Answer: $F \approx 0.040 N$ (lagbhag $4$ grams-force), attractive — balloon wall se chipak jaata hai.

Verify: $\approx 0.040 N$ ; yeh jo weight support kar sakta hai $= F / g = 0.040/9.8 \approx 0.0041 kg = 4.1 g$ ✓ — balloon kuch grams ka hota hai, isliye yeh plausibly chipakta hai. Realistic.

Cell I — exam twist (gravity se compare)

Worked example Example 9 — do protons ke liye electric vs gravity

Do protons ke liye ( $q = 1.6 \times 1 0^{- 19} C$ , $m = 1.67 \times 1 0^{- 27} kg$ ), unke electric repulsion aur gravitational attraction ka ratio nikalo. ( $G = 6.67 \times 1 0^{- 11}$ .)

Forecast: parent note ne do electrons ke liye $\sim 1 0^{42}$ find kiya. Protons bhaare hote hain, isliye gravity thodi zyada strong hai — kya ratio electrons ke liye se bada ya chhota hoga? (Guess: chhota, kyunki bhaara matlab zyada gravity.)

Step 1 — Dono forces likho aur ratio lo. $\frac{F _{e}}{F _{g}} = \frac{k q ^{2} / r ^{2}}{G m ^{2} / r ^{2}} = \frac{k q ^{2}}{G m ^{2}} .$ Ye step kyun? Dono forces identical $1/ r^{2}$ carry karte hain; ratio lene se $r$ completely cancel ho jaata hai — yahi twist ka poora trick hai. Answer ek pure number hai, separation se independent. (Dekho Newton's Law of Gravitation.)

Step 2 — Numbers daalo. $\frac{F _{e}}{F _{g}} = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{( 6.67 \times 1 0 ^{- 11} ) ( 1.67 \times 1 0 ^{- 27} ) ^{2}} .$ Top aur bottom alag kyun rakho? Har side mein ten ke bahut alag powers ka product hai; numerator poori tarah evaluate karo, phir denominator, phir ek baar divide karo — exponent bookkeeping honest rehti hai.

Step 3 — Top aur bottom evaluate karo. Top: $8.99 \times 1 0^{9} \times 2.56 \times 1 0^{- 38} = 2.30 \times 1 0^{- 28}$ . Bottom: $6.67 \times 1 0^{- 11} \times 2.79 \times 1 0^{- 54} = 1.86 \times 1 0^{- 64}$ . Ratio: $2.30 \times 1 0^{- 28} /1.86 \times 1 0^{- 64} = 1.24 \times 1 0^{36}$ .

Answer: $\frac{F _{e}}{F _{g}} \approx 1.2 \times 1 0^{36}$ .

Verify: Exponent $36 < 42$ (electron value) ✓ — protons sach mein bhaare hain, isliye gravity thodi gap close karti hai. Forecast confirm. Distance kabhi appear nahi hua: ratio $1 fm$ par ya ek galaxy ke across same hai.

Recall Yeh kaun sa cell hai? (quick self-test)

Ek charge teen doosron ke saath square ke corner par baitha hai. Kaun sa method? ::: 2-D vector addition (Cell F/F′): components, $x$ aur $y$ alag alag add karo, recombine karo — har charge ka sign push vs pull ke liye dekho. Bataya gaya hai ki ek charge $0$ hai. Force kya hai? ::: Exactly zero, kisi bhi distance par (Cell D). Do charges ek hi point par place kiye gaye, $r = 0$ . Formula kya deta hai? ::: Undefined — $r = 0$ forbidden hai; $F \to \infty$ jaise $r \to 0$ (Cell D′). "Net force zero kahan hai?" ::: Equilibrium point (Cell G): magnitudes equal set karo, square-root lo, solve karo. Distance triple, charge double — $F$ par factor? ::: $\times 2/9$ (Cell C): distance factor squared hota hai.

Active-Recall Flashcards

Coulomb force ki direction (attract/repel) kya set karta hai?

Product

q_{1} q_{2}

ka sign: positive → repel, negative → attract. Magnitude

∣ q_{1} q_{2} ∣

use karta hai.

Coulomb's law mein

r = 0

forbidden kyun hai?

Denominator mein

r^{2}

F

ko undefined bana deta hai (zero se division); jaise

r \to 0

F \to \infty

. Point charges coincide nahi kar sakte.

Teen charges ek line mein — ek par net force kaise nikalte hain?

Superposition: har pairwise force compute karo, use line ke along uski direction ke liye sign do, signed values add karo.

Non-collinear charges — method?

Har force ko

x

aur

y

components mein split karo, components alag alag add karo, recombine karo; har force ki direction us pair ke sign par depend karti hai (like ho toh push, unlike ho toh pull).

Do like charges ke beech equilibrium point ki condition?

k q_{1} / x^{2} = k q_{2} / (1 - x)^{2}

(test charge aur

k

cancel ho jaate hain); square-root lo aur solve karo.

Electric-to-gravity ratio mein

r

cancel kyun ho jaata hai?

Dono

F_{e}

aur

F_{g}

1/ r^{2}

carry karte hain; divide karne se woh remove ho jaata hai, ek pure constant bachta hai.